Problem: In the diagram, point $E$ lies on line segment $AB$, and triangles $AED$ and $BEC$ are isosceles. Also, $\angle DEC$ is twice $\angle ADE$. What is the measure of $\angle EBC$ in degrees? [asy]
import olympiad;
import math;

size(7cm);

// Draw triangles
pair a = (-2 * Cos(70), 0);
pair b = (1, 0);
pair c = dir(30);
pair d = dir(110);
pair e = (0, 0);
draw(a--d--e--cycle);
draw(e--b--c--cycle);

// Labels
label("$A$", a, W); label("$B$", b, E);
label("$C$", c, N); label("$D$", d, N);
label("$E$", e, S); label("$70^\circ$", a, 2 * NE+ 0.5 * E);

// Ticks
add(pathticks(a--d, s=2));
add(pathticks(d--e, s=2));
add(pathticks(e--c, 2, spacing=0.7, s=2));
add(pathticks(e--b, 2, spacing=0.7, s=2));
[/asy]
Explanation: Since $\triangle ADE$ is isosceles, then $\angle AED=\angle EAD=70^\circ$.

Since the angles in $\triangle ADE$ add to $180^\circ$, then $\angle ADE = 180^\circ - 2(70^\circ) = 40^\circ$.

Since $\angle DEC=2(\angle ADE)$, then $\angle DEC = 2(40^\circ)=80^\circ$.

Since $AEB$ is a straight line, then $\angle CEB = 180^\circ - 80^\circ - 70^\circ = 30^\circ$.

Since $\triangle EBC$ is isosceles, then $\angle ECB=\angle EBC$.

Thus, in $\triangle EBC$, $30^\circ + 2(\angle EBC)=180^\circ$ or $2(\angle EBC)=150^\circ$ or $\angle EBC=\boxed{75^\circ}$.